Question: Find the solution of the following equation whose argument is strictly between $90^\circ$ and $180^\circ$. Round your answer to the nearest thousandth. $z^2=-81i$ $z$ =
Answer: The Strategy A straightforward way to solve an equation of the form $z^{n}=m$ is by using the polar form of $z$. Therefore, our solution will consist of the following steps: Rewrite $z^n$ and $m$ in polar form. [How is this done, in general?] Solve for the modulus and argument of $z$. Find the rectangular form of $z$. [How is this done, in general?] Rewrite the equation in polar form Let's denote $r$ and $\theta$ to be the modulus and argument of $z$, respectively. Therefore, $z^{2}=r^{2}[\cos {({2}\cdot\theta)}+i\sin {({2}\cdot\theta)}]$. The number $-81i$ has a modulus of $81$. The argument of $-81i$ can be $270^\circ$ plus any multiple of $360^\circ$, so we can write it as $270^\circ+k\cdot360^\circ$ for an integer $k$. Now the equation looks as follows: $\begin{aligned}r^{2}[\cos {({2}\cdot\theta)}+i\sin {({2}\cdot\theta)}]&= \\81&[\cos(270^\circ+k\cdot360^\circ)+i\sin(270^\circ+k\cdot360^\circ)]]\end{aligned}$ When two complex numbers are equal, we know that both their moduli and arguments are equal. Therefore, we have the following equations for $r$ and $\theta$ : $r^{2}=81$ ${2}\cdot\theta=270^\circ+k\cdot360^\circ$ Solving for $r$ $\begin{aligned}r^{2}&=81 \\\\ r &=9 \end{aligned}$ Solving for $\theta$ $\begin{aligned}{2}\cdot\theta&=270^\circ+k\cdot360^\circ \\\\\theta&=135^\circ+k\cdot180^\circ\end{aligned}$ Remember that $\theta$ is strictly between $90^\circ$ and $180^\circ$. Therefore, we need to find the multiple of $180^\circ$ that is strictly within the range of $90^\circ-135^\circ=-45^\circ$ and $180^\circ-135^\circ=45^\circ$. This multiple is simply $0^\circ$, so $\theta=135^\circ$. Finding the rectangular form of $z$ Let's plug in $r=9$ and $\theta=135^\circ$ into the polar form of $z$ : $\begin{aligned}z&=r[\cos(\theta)+i\cdot\sin(\theta)]\\\\ &=9[\cos(135^\circ)+i\cdot\sin(135^\circ)]\\\\ &=9\cos(135^\circ)+9\sin(135^\circ)\cdot i\end{aligned}$ Using the calculator and rounding to the nearest thousandth, we get the following solution: $z=-6.364+6.364i$ Summary $z=-6.364+6.364i$